I need the work shown for solving the problems. Just want to make sure my answer
ID: 3260458 • Letter: I
Question
I need the work shown for solving the problems. Just want to make sure my answers are correct. I also use a TI-84 Plus calculator if a formula is needed, Thanks
6) For the quantitative variable, find what percentage of your data values fall within 1 standard deviation of your sample mean?; 2 standard deviations? How do your results compare to the empirical rule? 7) For the quantitative variable, approximate the top and bottom 5% cutoff points using the normal distribution. 8) Compute the standard error and the degrees of freedom for each set of data values. 9) Standardize the first score in your quantitative data set. 10) Select a point estimate for the population mean of the quantitative yariable and the population proportion ofthe qualitative variable. Create a 95% confidence interval about both of those estimates. 11) List how many data values you would need for an error half the size of the one you obtained for your confidence intervals in question no.10. Hours Worked Each week / Per Person (27) Color Votes Yes 19 0 24 40 40 20 40 0 35 40 4 27 0 20 27 25 32 25 30 0 0 27 30 140 30 30 0 25 Quantitative: How many hours do you work a week? Qualitative: Do you like the color yellow?Explanation / Answer
Solution
Quantitative Data – % within 1SD and 2 SD from the mean
Although all calculations can be done using the data as given using Excel Function, the calculations are done by the conventional method.
First step would be to obtain the frequency distribution and then compute mean and standard deviation using the formulae, mean (Xbar) = [[over all values of x](x.f)]/ f;
SD = sq.rt[[over all values of x]{(x – Xbar)2.f}/f], where f is the frequency of value x.
All steps are shown in the table below:
x
f
x.f
(x - Xbar)^2
(x - Xbar)^2.f
0
6
0
693.4426889
4160.656133
4
1
4
498.7762889
498.7762889
20
2
40
40.11068889
80.22137778
24
1
24
5.44428889
5.44428889
25
3
75
1.77768889
5.33306667
27
3
81
0.44448889
1.33346667
30
4
120
13.44468889
53.77875556
32
1
32
32.11148889
32.11148889
35
1
35
75.11168889
75.11168889
40
4
160
186.7786889
747.1147556
140
1
140
12920.11869
12920.11869
T0tal
27
711
18580
Xbar =
711/27 =
26.33333333
SD^2 =
18580/27 =
688.148148
Xbar to 4 decimal places
26.3333
SD =
sq.rt(688.148148)
26.232578
to 4 decimal
26.2325
Xbar – 1SD = 0; Xbar + 1SD = 52 => percentage within 1SD from mean = 26/27 = 0.9630 or 96.3%
Xbar – 2SD = - 26; Xbar + 2SD = 78 => percentage within 1SD from mean = 26/27 = 0.9630 or 96.3%
Comparison with Empirical Rule
For Normal Distribution,
P{( µ - ) X ( µ + )} = 0.68;
P{( µ - 2) X ( µ + 2)} = 0.95;
P{( µ - 3) X ( µ + 3)} = 0.997
i.e., percentage within 1SD from mean = 68 and percentage within 2SD from mean = 95.
While the first (96.3%) does not compare well with empirical rule, the second apparently compares well with the empirical rule.
[note: the above aberration is primarily because one data value, 140 is an out-lier. The word apparently has been used earlier because it so happens – generally with an out-lier there would be disparity there too.]
x
f
x.f
(x - Xbar)^2
(x - Xbar)^2.f
0
6
0
693.4426889
4160.656133
4
1
4
498.7762889
498.7762889
20
2
40
40.11068889
80.22137778
24
1
24
5.44428889
5.44428889
25
3
75
1.77768889
5.33306667
27
3
81
0.44448889
1.33346667
30
4
120
13.44468889
53.77875556
32
1
32
32.11148889
32.11148889
35
1
35
75.11168889
75.11168889
40
4
160
186.7786889
747.1147556
140
1
140
12920.11869
12920.11869
T0tal
27
711
18580
Xbar =
711/27 =
26.33333333
SD^2 =
18580/27 =
688.148148
Xbar to 4 decimal places
26.3333
SD =
sq.rt(688.148148)
26.232578
to 4 decimal
26.2325
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.